A metal sphere cools from `72^(@)C` to `60^(@)C` in 10 minutes. If the surroundings temperature is `36^(@)C`, then the time taken by it to cool from `60^(@)C` to `52^(@)C` is
A. 8 minutes
B. 4 minutes
C. 12 minutes
D. 10 minutes

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1 Answers

Correct Answer - D
`(d theta)/(dt)=k(theta_(AV)-theta_(0))`
`(12)/(10)=k(66-36)=k30`
`k=(12)/(300)=(1)/(25)`
`((d theta)/(dt))=k (theta_(AV)-theta_0)`
`(8)/(dt)=(1)/(25)(56-36)`
dt=10 min.

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